Consider the system of linear ODE's and its linear adjoint system given by \eqref{1} and \eqref{2} respectively ($x\in \mathbb{R}^n)$. $$ \begin{align} &x'(t)=M(t)x(t) \label{1} \tag{1}\\ &y'(t)=-M^\ast(t)y(t) \label{2}\tag{2} \end{align} $$ where $M$ is a continuous $n\times n$ matrix and $\omega>0$ periodic.

If we denote say $S$ and $S^*$ to be the space of $\omega$-periodic solutions to \eqref{1} and \eqref{2} respectively, is it true that their dimensions are equal?

I was thinking if we can show an element of $S^*$ is in the dual of $S$ and by using the fact that the dimension of a vector space and its dual must be equal, the proof then follows trivially.

I am not too sure how to solve this. Any help would be much appreciated. Thank you!


1 Answer 1


You can reformulate the problem into a purely algebraic one. Consider the flows of the two fields: $$ g'(t) = Mg(t) \\ \hat g'(t) = -M^\dagger \hat g(t) $$ with $g,\hat g$ invertible matrices. You can check that: $$ \hat g = (g^\dagger)^{-1} $$ (actually this is how you motivate the definition of the adjoint flow). Now consider $G = g(\omega)$, $\hat G = g(\omega)$. Your question question is therefore given an invertible matrix $G$, the fixed space of $G$, $F_G:= \{x\in\mathbb R^n|Gx = x\}$ and the fixed space of $\hat G := (G^\dagger)^{-1}$, $F_{\hat G}:= \{x\in\mathbb R^n|\hat Gx = x\}$ have the same dimensions: $$ \dim F_{\hat G} = \dim \hat F_G $$

Indeed, the evaluation at the period $\omega$ gives a natural isomorphism between the space of periodic solutions and the fixed space. It is well defined from the unicity of the solution given the initial condition.

Note that duality does not immediately apply. Indeed, the comparison would be direct if you wanted to relate the right fixed space of $G$, $S$, and the left fixed space of $\hat G$, $F_{\hat G}^L := \{x\in\mathbb R|x\hat G = x\}$. In this case, as the notation suggests, $F_{\hat G}^L$ is obtained by duality: $F_G^\dagger = F_{\hat G}^L$ so the spaces have the same dimension. However, you are rather interested in the right fixed space for both, so the easy argument does not apply.

First off, by inversion, $F_G = F_{G^{-1}}$, in particular they have the same dimensions. Thus you just need to check that $\dim F_G = \dim F_{G^\dagger}$. One way to show this is to use $\dim F_G = \dim Ker(G-I_n)= n-\text{rank}(G-I_n)$, $(G-I_n)^\dagger = G^\dagger-I_n$, and the rank is preserved by the adjoint. This allows you conclude.

I don't think that there is a natural isomorphism between the two spaces. Indeed, it amounts to constructing an isomorphism between left and right eigenspaces, which is not possible without additional structures.

Hope this helps.

  • $\begingroup$ Thankk. But can you please think about this way: if x is the w periodic solution, then for prcipal fundamental matrix we have $x(\omega)=X(\omega)x(0)$. Then $x(0)\in kernel(X(\omega)-I)$. We want to show that the kernel space have same dimension. I am stuck here. $\endgroup$
    – Hermi
    Oct 20, 2023 at 3:10
  • $\begingroup$ That’s the method I used. I proved in the second to last paragraph… $\endgroup$
    – LPZ
    Oct 20, 2023 at 6:23
  • $\begingroup$ I agree with rank preserve the adjoint. But your matrix G of adjoint system has a inverse, right?Is it still same rank as G? $\endgroup$
    – Hermi
    Oct 20, 2023 at 15:37
  • $\begingroup$ You need to apply the reasoning to $G^{-1}$. You therefore have $\dim\text{Ker}(G-I)= \dim\text{Ker}(G^{-1}-I)= \dim\text{Ker}((G^{-1})^\dagger-I)$ $\endgroup$
    – LPZ
    Oct 20, 2023 at 18:27

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